def bubble_down(values: list, left: int, right: int) -> None:"""Bubble down through values[left: right+1], swapping items that are outof order. Note that use of this slicing notation means that itemsvalues[left], values[left + 1], values[left + 2],be modified.values[right] couldPrecondition: left and right are valid indexes in values.>> list_example_1>> bubble_down(list_example_1, 1, 3)>> list_example_1[4, 1, 3, 2, 0]>> list_example_2 = [4, 3, 2, 1, 0]>> bubble_down(list_example_2, 0, 4)>> list_example_2[о, 4, 3, 2, 1][4, 3, 2, 1, 0]

According to the information given:-We have to write the bubble_down function to swap the items in…

def bubble_down(values: list, left: int, right: int) -> None: """Bubble down through values[left: right+1], swapping items that are out of order. Note that use of this slicing notation means that items values[left], values[left + 1], values[left + 2], II I values[right] could be modified. Precondition: left and right are valid indexes in values. [4, 3, 2, 1, 0] >> list_example_1 >> bubble_down(list_example_1, 1, 3) >> list_example_1 [4, 1, 3, 2, 0] >> list_example_2 >> bubble_down(list_example_2, 0, 4) >> list_example_2 [о, 4, 3, 2, 1] [4, 3, 2, 1, 0]

def bubble_down(values: list, left: int, right: int) -> None: """Bubble down through values[left: right+1], swapping items that are out of order. Note that use of this slicing notation means that items values[left], values[left + 1], values[left + 2], II I values[right] could be modified. Precondition: left and right are valid indexes in values. [4, 3, 2, 1, 0] >> list_example_1 >> bubble_down(list_example_1, 1, 3) >> list_example_1 [4, 1, 3, 2, 0] >> list_example_2 >> bubble_down(list_example_2, 0, 4) >> list_example_2 [о, 4, 3, 2, 1] [4, 3, 2, 1, 0]

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Consider the following linked list of the form. This list already exists, so you do not need to write the code to create the list below: struct nodeType { int infoData; nodeType * next; }; nodeType *first; … and containing the values: first (( See image for values )) Use this linked list to perform the following actions. Declare any additional pointers, etc., if needed, to complete the code. You do not need to write any includes or using statements—just program segments. Write a code segment that creates a new node with infoData = 13 and inserts the node after the node with the value 12.
True/False Select true or false for the statements below. Explain your answers if you like to receive partial credit. 3) Which of the following is true about the insertBeforeCurrent function for a CircularLinked List (CLL) like you did in programming exercise 1?a. If the CLL is empty, you need to create the new node, set it to current, andhave its next pointer refer to itselfb. The worst case performance of the function is O(n)c. If you insert a new element with the same data value as the current node, theperformance improves to O(log n)
def bubble_down(values: list, left: int, right: int) -> None: """Bubble down through values[left: right+1], swapping items that are out of order. Note that use of this slicing notation means that items values[left], values[left + 1], values[left + 2], be modified. values[right] could Precondition: left and right are valid indexes in values. >> list_example_1 >> bubble_down(list_example_1, 1, 3) >> list_example_1 [4, 1, 3, 2, 0] >>> list_example_2 >> bubble_down(list_example_2, 0, 4) >>> list_example_2 Со, 4, 3, 2, 1] [4, 3, 2, 1, 0] [4, 3, 2, 1, 0] II II ||
PYTHON: Please fix my code to print the correct traversals, especially at the def functions, because I could not print the correct results of the inorder, postorder, and preorder traversals from the user input. #POSTORDERclass Node: def __init__(self, data): self.left = None self.right = None self.data = data # Insert Node def insert(self, data): if self.data: if data < self.data: if self.left is None: self.left = Node(data) else: self.left.insert(data) elif data > self.data: if self.right is None: self.right = Node(data) else: self.right.insert(data) else: self.data = data # Print the Tree def PrintTree(self): if self.left: self.left.PrintTree() print(self.data),…
Having trouble with creating the InsertAtEnd function in the ItemNode.h file below. " // TODO: Define InsertAtEnd() function that inserts a node // to the end of the linked list" Given main(), define an InsertAtEnd() member function in the ItemNode class that adds an element to the end of a linked list. DO NOT print the dummy head node. Ex. if the input is: 4 Kale Lettuce Carrots Peanuts ------------------------------------------------------- main.cpp -------------------------------------------------------- #include "ItemNode.h" int main() { ItemNode *headNode; // Create intNode objects ItemNode *currNode; ItemNode *lastNode; string item; int i; int input; // Front of nodes list headNode = new ItemNode(); lastNode = headNode; cin >> input; for (i = 0; i < input; i++) { cin >> item;…
The following code doesn't correctly return the index values of the minimum contiguous subsequence. Please modify it so that it does, with comments if possible. def minSumRec(a, left, right): if left == right: return [a[left], left, right] center = int((left + right) / 2) min_left_sum = minSumRec(a, left, center) min_right_sum = minSumRec(a, center + 1, right) min_left_border_sum, left_border_sum = 1e9, 0 for i in range(center, left - 1, -1): left_border_sum += a[i] if left_border_sum < min_left_border_sum: min_left_border_sum = left_border_sum min_right_border_sum, right_border_sum = 1e9, 0 for i in range(center + 1, right + 1): right_border_sum += a[i] if right_border_sum < min_right_border_sum: min_right_border_sum = right_border_sum min_border_sum = min_right_border_sum + min_left_border_sum minSum = min(min_left_sum[0], min_right_sum[0], min_border_sum) if minSum == min_left_sum: return min_left_sum if minSum == min_right_sum: return min_right_sum return…
def bubble_up(values: list, left: int, right: int) -> None: """Bubble up through values[left: right+1], swapping items that are out of order. Note that use of this slicing notation means that items values[left], values[left + 1], values[left + 2], ..., values[right] could be modified. Precondition: left and right are valid indexes in values. >>> list_example_1 = [4, 3, 2, 1, 0] >>> bubble_up(list_example_1, 0, 3) >>> list_example_1 [3, 2, 1, 4, 0] >>> list_example_2 = [4, 3, 2, 1, 0] >>> bubble_up(list_example_2, 2, 4) >>> list_example_2 [4, 3, 1, 0, 2] """ for i in range(left, right): if values[i] > values[i + 1]: temp = values[i] values[i] = values[i + 1] values[i + 1] = temp # Use the bubble_up function as a starting point to help you complete the# bubble_down function.def bubble_down(values: list, left: int, right: int) -> None:…
You will reverse your student ID by the linked list of Struct in C!!!! Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Create a linked list and insert your student ID elementwise to the list. Reverse the linked list and printout the result. Hint: You can use additional pointers to reverse the linked list. You can also use a stack and then simply assign the element to the stack and it will be reversed. Linked list, Stack, Queue example source code is available on week 7 resources section.
def remove_after_kth(items, k=1): Given a list of items, some of which may be duplicated, create and return a new list that is otherwise the same as items, but only up to k occurrences of each element are kept, and all occurrences of that element after those first k are discarded. Hint: loop through the items, maintaining a dictionary that remembers how many times you have already seen each element. Update this count as you go, and append each element to the resultlist only if its count is still at most equal to k. items k expected results [42, 42, 42, 42, 42, 42, 42] 3 [42, 42, 42] ['tom', 42, 'bob', 'bob', 99, 'bob', 'tom', 'tom', 99] 2 ['tom', 42, 'bob', 'bob', 99, 'tom', 99] [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1] 1 [1, 2, 3, 4, 5] [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5] 3 [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 1, 5] [42, 42, 42, 99, 99, 17] 0 [ ] Please add comments between the codes so I can understand
def update_messages (msgs: List [str], lens: List [int]) -> None: ""Update the list of the strings in msgs so their lengths are adjusted to match the value at the corresponding position in lens. Strings that are too long should be shortened by leaving off characters at the end and strings that are too short should be lengthened by adding underscores to the end. Precondition: len (msgs) == len(lens) and all values in lens are >= 0 ['aardvark', 'bear', 'cat', 'dog'] >>> update_messages (messages, [6, 5, 5, 3]) >>> messages = >>> messages ['aardva', 'bear_', 'cat__', 'dog'] >>> messages = ['', 'cat'] >> update_messages (messages, [3, 0]) >>> messages ['---', '']
Please write your own code, I'm not understanding
Do not add import statements